Mass budgets — dry mass, wet mass, margin: Exercises

This page is a self-test. Read each problem, try it yourself, then open the collapsible Solution to check every step. Problems climb from L1 (just recognise the words) to L5 (design a full budget under conflicting constraints). If a symbol looks unfamiliar, revisit the parent note first.

Everything here rests on three ideas built in the parent:

Recall The three quantities you need (open if rusty)

Dry mass $m_{dry}$ ::: everything except consumables — structure, payload, empty tanks, wiring.
Wet mass $m_{wet}$ ::: dry mass plus propellant (and pressurant/other consumables).
Mass margin ::: a reserved fraction $ϵ$ of the estimate held back as insurance against growth: $m_{margin} = ϵ m_{dry,est}$ . Here $m_{dry,est}$ is exactly the CBE (Current Best Estimate) — the raw sum of component masses with no cushion added. So "estimate" and "CBE" mean the same number on this page.

And the one equation that ties mass to motion — the Tsiolkovsky Rocket Equation: $Δ V = v_{e} ln (\frac{m _{wet}}{m _{dry}})$ where $v_{e}$ is the exhaust velocity (how fast gas leaves the nozzle, in $m/s$ ), tied to Specific Impulse by $v_{e} = I_{s p} g_{0}$ with $g_{0} = 9.81 m/s^{2}$ .

Intuition The fine print behind every formula on this page

Tsiolkovsky assumes three things, and all our answers inherit them:

Constant $v_{e}$ — the engine's exhaust speed doesn't change mid-burn. Real engines drift a little; we treat it as fixed.
No gravity or drag losses — the equation gives ideal $Δ V$ (the velocity change if the ship coasted in empty space). Real maneuvers near a planet lose some $Δ V$ to gravity and air, so the mission $Δ V$ you plug in is usually padded to cover those losses (see Mission ΔV Budget).
All propellant leaves as exhaust — no leaks, no unusable residuals (in practice a few percent is trapped).

Edge cases to keep in mind:

$Δ V \to 0$ : then $e^{Δ V / v_{e}} \to e^{0} = 1$ , so $m_{wet} \to m_{dry}$ — zero maneuver needs zero propellant. Good sanity check.
$ϵ = 0$ (no margin): $m_{margin} = 0$ , so the design carries no cushion — the moment any component grows, the budget breaks. This is why real programs never fly $ϵ = 0$ until hardware is built and weighed.
$v_{e} \to \infty$ (a perfect, infinitely efficient engine): $Δ V / v_{e} \to 0$ , so again $m_{wet} \to m_{dry}$ — you'd need almost no propellant. This is why high Specific Impulse is prized.

Level 1 — Recognition

Goal: identify which bucket a mass goes into and read the definitions correctly.

The figure below is the mental picture behind every problem on this page — where each kilogram lives:

Figure — Mass budgets — dry mass, wet mass, margin

Recall Solution 1.1

What we do: sort each item into dry vs. wet. Dry = everything that is not consumable. The empty tank counts as dry (it's hardware); the hydrazine is the only consumable. $m_{dry} = 120 + 40 + 25 + 15 = 200 kg$ $m_{wet} = m_{dry} + m_{prop} = 200 + 300 = 500 kg$ Answer: dry $= 200 kg$ , wet $= 500 kg$ .

Recall Solution 1.2

CBE (Current Best Estimate) is the raw sum with no margin. MEV adds the margin: $MEV = CBE \times (1 + ϵ) = 800 \times 1.25 = 1000 kg$ Answer: $1000 kg$ .

Level 2 — Application

Goal: plug numbers into the rocket equation, both directions.

Recall Solution 2.1

Why the exponential form? We know the dry (final) mass and want the wet (initial) mass, so we rearrange Tsiolkovsky to solve for $m_{wet}$ . Undo the $ln$ by exponentiating: $m_{wet} = m_{dry} e^{Δ V / v_{e}} = 200 e^{1.2/3.0} = 200 \times e^{0.4}$ Here both $Δ V$ and $v_{e}$ are already in the same unit (km/s), so the ratio $1.2/3.0 = 0.4$ is a clean pure number — no conversion needed. $= 200 \times 1.4918 = 298.4 kg$ $m_{prop} = m_{wet} - m_{dry} = 298.4 - 200 = 98.4 kg$ Answer: propellant $\approx 98.4 kg$ , wet mass $\approx 298.4 kg$ .

Recall Solution 2.2

Step 1 — get $v_{e}$ . Specific Impulse $I_{s p}$ (in seconds) converts to exhaust velocity by multiplying by $g_{0} = 9.81 m/s^{2}$ : $v_{e} = I_{s p} g_{0} = 220 \times 9.81 = 2158.2 m/s$ Notice this comes out in m/s. Step 2 — match the units before dividing. The exponent $Δ V / v_{e}$ must be a pure number, so both quantities have to be in the same unit. Our $v_{e}$ is in m/s, so we convert $Δ V$ to m/s too: $Δ V = 0.5 km/s = 0.5 \times 1000 = 500 m/s$ Why this step matters: if we left $Δ V$ as $0.5 km/s$ over $2158.2 m/s$ we'd compute $0.5/2158.2 = 0.00023$ instead of $500/2158.2 = 0.2317$ — off by a factor of 1000 and a nonsense answer. Step 3 — propellant. $m_{wet} = 50 e^{500/2158.2} = 50 e^{0.2317} = 50 \times 1.2607 = 63.03 kg$ $m_{prop} = 63.03 - 50 = 13.03 kg$ Answer: $v_{e} \approx 2158 m/s$ , propellant $\approx 13.0 kg$ .

Level 3 — Analysis

Goal: track how a change (mass growth, a burn) propagates, and interpret the number.

Recall Solution 3.1

Why this formula? From the parent derivation, holding $Δ V$ fixed while dry mass grows forces wet mass up in proportion, and the extra propellant is the dry growth amplified by the penalty factor: $Δ m_{prop} = Δ m_{dry} (e^{Δ V / v_{e}} - 1)$ Compute the penalty factor first (both speeds in km/s, so $6/3 = 2$ is a clean pure number): $e^{6/3} - 1 = e^{2} - 1 = 7.389 - 1 = 6.389$ $Δ m_{prop} = 8 \times 6.389 = 51.1 kg$ Interpretation: each kilogram of structure growth costs $\approx 6.4 kg$ of propellant — the whole reason margins are guarded. The red curve in the figure below shows how violently this penalty factor climbs as $Δ V / v_{e}$ grows.

Figure (above): horizontal axis is the dimensionless ratio $Δ V / v_{e}$ ; vertical axis is how many kilograms of propellant each extra kilogram of dry mass forces onboard. The black dots mark a GEO-class maneuver ( $Δ V / v_{e} = 0.5$ , penalty $\approx 0.65$ ) and this Mars case ( $Δ V / v_{e} = 2$ , penalty $\approx 6.4$ ). Notice the curve is nearly flat at the left and rockets upward past $Δ V / v_{e} \approx 1$ — that steepening is the exponential punishing high- $Δ V$ missions.

Answer: $\approx 51.1 kg$ of extra propellant.

Recall Solution 3.2

Why divide, not multiply? In L2 we sized a tank: we knew the final (dry) mass and asked how big the ship must be before the burn, so we multiplied up to the heavier initial mass. Here it is the opposite question — we already know the initial (heavier) mass and ask what remains after fuel is spent. Start from Tsiolkovsky $Δ V = v_{e} ln (m_{i} / m_{f})$ and solve for $m_{f}$ : exponentiate to get $m_{i} / m_{f} = e^{Δ V / v_{e}}$ , then flip it, $m_{f} = m_{i} / e^{Δ V / v_{e}}$ . The final (lighter) mass sits in the denominator, so we divide the known initial mass by the ratio $R$ . Dividing makes the ship lighter — exactly what burning fuel should do. $mass ratio R = e^{3800/4414} = e^{0.8609} = 2.3654$ (both speeds already in the same km/s unit, so $3800/4414$ works whether we write them as m/s or km/s — the ratio is identical.) $m_{after} = \frac{45000}{2.3654} = 19, 024 kg$ $m_{prop,burned} = 45000 - 19024 = 25, 976 kg$ $m_{prop,remaining} = m_{after} - m_{dry} = 19024 - 5000 = 14, 024 kg$ Answer: after burn $\approx 19, 024 kg$ ; burned $\approx 25, 976 kg$ ; remaining propellant $\approx 14, 024 kg$ .

Level 4 — Synthesis

Goal: combine margin bookkeeping with the rocket equation into one budget.

The figure traces how one kilogram of raw component mass swells on its way to the launch pad — margin first, then the rocket-equation mass ratio:

Recall Solution 4.1

Step 1 — dry CBE. Why: the CBE is our raw best estimate, the starting point before any cushion; we add the component masses. $CBE = 1.2 + 0.8 + 1.0 = 3.0 kg$ Step 2 — apply dry margin. Why: real hardware grows (adhesive, harness, coating), so we inflate the estimate by $ϵ = 0.20$ to get the design dry mass we actually build the tank around. $m_{dry} = 3.0 \times 1.20 = 3.6 kg$ Step 3 — convert $I_{s p}$ to $v_{e}$ . Why: the rocket equation needs a velocity, not seconds; multiply by $g_{0}$ . $v_{e} = 65 \times 9.81 = 637.65 m/s$ Here $Δ V = 50 m/s$ is already in m/s and matches $v_{e}$ 's unit, so no conversion is needed. Step 4 — propellant (size up). Why: we know the design dry (final) mass and want the wet (initial) mass, so we multiply up by the mass ratio, just like L2. $m_{wet} = 3.6 e^{50/637.65} = 3.6 e^{0.07841} = 3.6 \times 1.08157 = 3.8937 kg$ $m_{prop} = 3.8937 - 3.6 = 0.2937 kg$ Step 5 — propellant margin. Why: residuals, valve leakage and off-nominal burns eat propellant too, so we add a $10%$ reserve on the propellant itself. $m_{prop,tot} = 0.2937 \times 1.10 = 0.3231 kg$ Step 6 — final wet mass. Why: the deliverable is the total mass on the pad = design dry + total propellant. $m_{wet,final} = 3.6 + 0.3231 = 3.923 kg$ Check: $3.923 kg ≪ 12 kg$ ✓ — plenty of headroom (see Structural Mass Fraction for how much of that is frame). Answer: wet mass $\approx 3.92 kg$ ; fits comfortably.

Recall Solution 4.2

(a) Original MEV. Why: MEV is the CBE inflated by the margin fraction — the ceiling the program commits to. $MEV = 1200 \times 1.25 = 1500 kg$ (b) Remaining margin. Why: every kilogram of real growth eats into the fixed MEV ceiling. First sum the growth, add it to the old CBE to get the new CBE, then see how far the ceiling sits above it. $growth = 80 + 30 + 15 = 125 kg, new CBE = 1200 + 125 = 1325 kg$ $remaining margin = 1500 - 1325 = 175 kg$ (c) Margin percentage. Why: margin is meaningful only relative to what the thing currently weighs — our best current estimate, the new CBE — not the frozen ceiling. $\frac{175}{1325} = 0.1321 = 13.2%$ Interpretation: margin fell from $25%$ to $13.2%$ . Below a typical $\sim 15%$ PDR floor — time for a mass-reduction "tiger team." Answer: MEV $= 1500 kg$ ; remaining $175 kg$ ; $13.2%$ .

Level 5 — Mastery

Goal: design under a hard constraint — trade margin, $Δ V$ , and payload against each other.

Recall Solution 5.1

Step 1 — name the unknown. Why: payload is what we're solving for, so give it a symbol and express dry mass in terms of it. Let payload $= P$ , so total dry CBE $= 150 + P$ . Step 2 — apply margin. Why: the tank must be sized around the design dry mass, which is the CBE inflated by $20%$ : $m_{dry} = 1.20 (150 + P) .$ Step 3 — link to the launch cap via the rocket equation. Why: the wet mass the mission demands is the design dry mass multiplied up by the mass ratio (we know final, want initial — same "multiply up" as L2), and that wet mass cannot exceed what the launcher lifts: $m_{wet} = 1.20 (150 + P) e^{Δ V / v_{e}} \leq 600, e^{2/3} = 1.9477$ (both speeds in km/s, so $2/3$ is a clean pure number.) Step 4 — solve the inequality. Why: isolate $P$ by dividing out the two amplifying factors. $1.20 \times 1.9477 (150 + P) \leq 600 \Rightarrow 2.3372 (150 + P) \leq 600$ $150 + P \leq \frac{600}{2.3372} = 256.72 \Rightarrow P \leq 106.72 kg$ Answer: maximum payload $\approx 106.7 kg$ . Sanity: at $P = 106.7$ , dry CBE $= 256.7$ , design dry $= 308.1$ , wet $= 308.1 \times 1.9477 = 600$ ✓ — exactly at the cap.

Recall Solution 5.2

Step 1 — baseline wet mass ( $P = 90$ ). Why: we need the "before" number to measure the change against. Apply the same margin-then-mass-ratio chain as 5.1. $dry CBE = 240, m_{dry} = 1.20 \times 240 = 288, m_{wet} = 288 \times 1.9477 = 560.94 kg$ Step 2 — post-swap wet mass ( $P = 110$ ). Why: redo the identical chain with the heavier payload to get the "after" number. $dry CBE = 260, m_{dry} = 1.20 \times 260 = 312, m_{wet} = 312 \times 1.9477 = 607.68 kg$ (a) Rise in wet mass. Why: subtract before from after. $607.68 - 560.94 = 46.74 kg$ Notice the $20 kg$ payload swap costs $46.7 kg$ of wet mass — the $\times 1.20$ margin and the $\times 1.9477$ mass ratio both amplify it ( $20 \times 1.20 \times 1.9477 = 46.74$ ). (b) Slack under cap. Why: the cap minus the demanded wet mass tells you if you still fit. $600 - 607.68 = - 7.68 kg (negative \to over the launcher by 7.7 kg)$ (c) Consequence if slack is negative. You must recover mass elsewhere — reduce $Δ V$ (a smaller ΔV budget, meaning less orbit capability), lighten structure, or move to a bigger (costlier) launch vehicle. Answer: (a) $+ 46.7 kg$ ; (b) $- 7.7 kg$ (over by $7.7 kg$ ); (c) e.g. cut $Δ V$ / descope / upsize launcher.